Σ = 0 N + F Sin 37° = N + (20)(0,6) = 40 N + 12 = 40 N = 28 N Σ = . F cos 37° = . (20)(0,8) μN = 4. a 16 (0,4)(28) = 4 16 11,2 = 4 4,8 = = 1,2 4
= + = 0 + (1,2)(5) = 6 /
1.2 /
: = 80 / = 336.000 /
Σ = 0 F = 0 = = Σ = 0 = 0 =
= = ()
= ∆ ∙ = ∙ ∙ = ∙ ( ) ∙ = ∙ (0,4)(25)(10)(21) = (336000) 2100 = 336000 2100 = = 6,25 × 10− = 6,25 336000
F 0 F C W 0 F C w q. E m. g 8 10
E
6
E 2 10
6
10
20 8
2,5 N / C arahnya ke bawah
3 2
5 2 5 3
rad
v
f
30
5
6m
θ x
2
A B
3 2
3
B
2
6
4
3 4 3
30
A cos B
10. cos 30
10 .
2
6
B
B
y
4
1
3
2 5 3
cm
0
0
Δ
t .t 0
t 0
1
v
2
c
2
12
1
(0,6c) c
2
2
12 1 0,36
12 0,8
15 tahun
v B v B v B
2
2
2
v A
0 2 gh
2 gh
2 gh
y
1
v0 t gt 2 2
4 v B
0,2
1
2 4 0,2v B 0,2 0,2v B
v B
3,8
19
2 gh
19
2 gh 19
2
20h 361
h 18,05 m
10 0,2
2
ρ
P 1V 1
P 2V 2
P gh V P 3V 0
P 0
0
gh 3 P 0
gh
2 P 0
1000 10 h
h 20 m
2 1 10
5
1 2 1 8 3 8
I 0
I 0
I 0
1 32 3 32
I 0
I 0
Terlebih dahulu kita harus mencari besar intensitas cahaya yang keluar dari polarisator III (I 3) dengan sudut antara polarisator III dan polarisator I adalah 30 0.
I 3
I 1 cos 2
1 2
2
I 0 . cos 30
0
1 I 0 . 2 2 1
2
3
1 2
I 0 .
3 4
3 8
I 0
Nilai intensitas cahaya yang keluar dari polarisator III tersbut disubtitusikan ke persamaan yang sama, dengan sudut antara polarisator II dan III adalah : 900 - 300 = 600. Maka besar intensitas cahaya yang keluar dari analisator (I 2) adalah :
I 2
I 3 cos 2
3 8
2
I 0 . cos 60
0
2
3 1 3 1 I 0 . I 0 . I 0 8 2 8 4 32 3
= ℎ ℎ
= , = , , = 0,5
= = × 0,5 = 0,75
= 22,4 = 0,75 22,4 = 16,8 = × 16,8 = 84
⇌
⇌
( )( ) = ( ) , = = , , = = , , = = ,
0,2 0,2 = 1,2 0,8 1,2 1,2 0,2 1,2 = 0,8 1,2 = 28,8
= ∆ = 0 ( ) = 0 = 0,87 ℃ = ∆ = 0 ( ) = 0 9 = 0,43 ℃ = ∆ = 0 ( ) = 0 = 0,67 ℃ = ∆ = 0 ( ) = 0 , 2 = 1,37 ℃ = ∆ = 0 ( ) = 0 3 = 1,13 ℃
= = 0,4 0,1 1 = 2 0,04 = 2 = 0,02 = 171 / = 3,42 0,02 = 1 2 2 = 216 21 171 = 34 = 171 34 =
ℎ] [+] = [ [] [+] = 1 10− 0,4 0,4 [+] = 1 10− = log[+] = log[1 10−] =5
lim f ( x) 3g ( x) 2
lim 3 f ( x) g ( x) 1
x a
lim f ( x) g ( x)
x a
x a
1 1
2
1
4 1 4 1 2
lim f ( x ) 3 g ( x) 2 3 lim 3 f ( x) 9 g ( x) 6 x a x a
lim 3 f ( x) g ( x) 1
1
x a
lim = →
lim 3 f ( x) g ( x) 1
lim (10) = 5 → lim = →
lim f ( x) g ( x) lim f ( x) lim g( x) x a
x a
−− = lim → −+− 1 12 1 8 1 12 1 8 1 4
x a
x a
1
1
2
1
2
4
1 3 1 1 2sin cos 3 1 2 lim = lim → 4 3sin3 9 → 3x 1sin 3 3 1 3.sin 1 3 2.sin 2 2 = lim → 3x 1sin 3 3 1.1 2. 22 = lim → x 1.3 1 = 2.32 = 121
x
8
4
4
tan x
i
i 0
4
sin x
i
i 0
sin x
8
2i
i 0
1 tan x i
8
2i
i 0
sin x
8
2i
i 0
f x
8 tan
'
S
2
'
2
2
8 sec x
x
1
1 r
1 1 sin 0 8 sin x
f ( x) 8 '
f ( x)
1 8 1 cos x 1 sin
f ( x) 8
f ( x )
'
x
2
8 1 sin x
2
4
sin x sin x sin x ... 2
4
x sin x sin x ... 6
6
'
f ( x ) 8 sin
2.0
x sin
2.1
x sin
2.2
2.3
x sin
x ... 8 sin x
2i
i 0
α
β
α β
α α β
α
β 4 5
β
2
2
5
2
2
6 4 5
24
19
β α
4 5
30 4
30
2
5
5 6 5 0 α
β
x
x
2 1
x
x
1 2 1
β
β
x
x
2 1
2
x
x
2 1
2
x
x
2 1
1
x
x
x
x
x
x
x
x
1 0 2 1
2 1
1 2
1
2
4
x
x
2
x
x
2 2
x
x
0 2 2
0
3
x
2
x
2
0
α
α
√ √ 7√ √ √
H
G
E
2
F
D
C 3
A
4
B
13
20
H
C
A
= 2.. 1 3 5 2 0 √ √ = 2. √ 13.5 25 20 = 13 10√ 13 = 1018√ 13 √ 13 = 95.13 = 9√ 6513 a
a
a b
a b
a
a
b
b
b
b
cos
a b ,
1 2 5
2
a b
5
a b
o
2 5 cos60
b
b
a
a b b a b
b
2
b
b
a b
b
2
b
5 25
6
5
b
3 80 6 80 16 80 40 80 46 80
8
8
82
10
2
2
82
10
5
5
53 3
8 3
53
P M A
8
P M P P P P B
A
A
5
10 8
2
3
10 8
40
6
80
8
46
80
80
1
1 x
y
y
5
z
x
5
y
2
z
2
xy xz
1
5
52
2
1 x y
2
2
y
z
2
5 2
52 1
1
x y y z
5
2
5
5
2
5
2
2
5
x z 2
5
2
= 2 2 2 2 2 2 √ 5 2 √ 5 2 2√ 5 = 2 2 2 2 2 2 5 4√ 5 4 5 4√ 5 4 20 = 2 2 2 2 2 2 38 = 2 2 2 2 2 2 19 =
4
3
3
4
4 3
3
4
2
3
yz
6 2 r
2 r
4s
6 4 s 6 2 r
s
4
= = 3 2 9 3 = 4 2 4 3 2 ′ = 2 2 4 = 0 2 3
2 4 = 2 2 2 = 32 2 2 = 32 4 2 = 32 = 4 3
3 r
2
2 + 1 = 2 + 1 = 21 + (1 1 ) = 1 21 + = 1
2 + 1 = × 2 + 21 = 1 × 1 4 + 21 = 2 + 21 = 1 − 3 = 2 1 3 = 2 2 + 1 3 = + 2 1 = + = + =2
●
●
= + ⇔ ′ = 1 = ′1 = 1 = 2
●
●
●
●
-
-
Sebagian besar murid pandai berenang
Sebagian besar murid pandai mendayung
+16
+8
+4
+8
+32
+32
+16
1
1 1 ;0,65;1 ; ;37,5%. 2 7 8 1
1 1 ;1 ; ;37,5%;0,65 2 7 8 1 1 1 ; ;37,5%;0,65;1 8 2 7 1 1 1 ;37,5%; ;0,65;1 8 2 7 1 1 1 ; ;0,65;37,5%;1 8 2 7 1 1 1 37,5%; ; ;0,65;1 2 8 7
1 8
0,125;37,5%
0,375;
1 2
0,5;0,65;1
1 7
9
7
1,286
20
1
bagian 120 6 20 2 B bagian 90 9 20 1 C bagian 60 3 A
1 6
a.
b.
c.
2 9
1 3
d.
13 18
0,72 bagian.
e.
a.
b.
e.
d.
c.
m c 8 2
42 cm
Luas bangunan = Luas 1/2 lingkaran + Luas trapesium
Jumlah sisi sejajar r 2 tinggi 2 22 42 28 142 28 7 2 616 980 1.596 cm2
1. 2.
3. 4. 5.
2 = 20 −
(1) (2) (3) (4)
-
-
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
